Question

Solve it.
With proper explanation.
Warnings : -
1. Irrelevant answers will be reported.
2. Answer with proper explanation.
​

Answer 1

Answer:

Formula for solving inclination of mirror type questions:

$\boxed{ \text{No. of Images} = \dfrac{360^\circ }{ \theta}}$

where, θ is the angle between the 2 mirrors.

If the number of images obtained is even, then '1' is subtracted to obtain the number of images.

This is because, the object lies in the angle bisector of the mirror due to which we get 1 extra image (which is the reflection of an image). Hence we subtract 1 to ignore that.

Q1) Angle = 45°.

Substituting in the formula we get:

⇒ 'n' = (360/45) - 1

⇒ n = 8 (even)

⇒ n = 8 - 1 = 7 images

Q2) Angle = 60°

⇒ n = (360/60) - 1

⇒ n = 6 (even)

⇒ n = 6 - 1 = 5 images

Q3) Angle = 120°

⇒ n = (360/120) - 1

⇒ n =  3 images (odd)

Q4) Mirrors are assumed to be adjacent sides of the room.

Hence the angle would be 90°

⇒ n = (360/90) - 1

⇒ n = 4 (even)

⇒ n = 4 - 1 = 3 images

Q5) Similar to last question, a new variation is that, a mirror on the floor is added. Let us first assume only adjacent sides of the mirror. Then the angle would be 90° and we get 3 images.

Now we have a mirror at the bottom as well. Hence these 3 images and the object will form a reflection on the mirror (floor). So the total number of images would be:

⇒ n = 3 (adjacent) + 3 (reflection on floor) + 1 (reflection of object)

⇒ n = 7 images

Q6)

Man Velocity with respect to ground = 5 m/s

Image Velocity with respect to ground = ?

We know that,

In a plane mirror, the object velocity = image velocity in terms of magnitude. Hence the image velocity would have a magnitude of 5. Since the direction of movement is opposite, we have an negative '-' sign.

Hence the velocity of image w.r.t ground is -5 m/s.

Q7)

⇒ Velocity of man = 10 m/s in the negative direction (away)

⇒ Velocity of man = -10 m/s

Therefore the velocity of image would be moving opposite to that of man. Hence it would have a '+' sign. Therefore velocity of image w.r.t ground is + 10 m/s. But the question has asked velocity of image w.r.t to man.

⇒ Velocity of image w.r.t man = Velocity of image - Velocity of man

⇒ Velocity of image w.r.t man = 10 - ( -10 ) = 20 m/s

Q8)

Velocity of mirror = 5 m/s towards the man.

Velocity of image = ?

Since mirror approaches the man with -5 m/s, the velocity of man w.r.t mirror is also 5 m/s. Therefore Velocity of image would also be 5 m/s but in the opposite direction. Hence, we get:

⇒ Velocity of image = 5m/s - ( -5 m/s) = 10 m/s

Q9) Since the mirror is moving away, the speed of the image would be same as that of mirror in the same direction of the mirror. But since the mirror is moving away from the object, it would have a negative sign.

⇒ Velocity of mirror = -6 m/s

⇒ Velocity of image w.r.t ground = -6 m/s

Q10)

Velocity of Man = 5 m/s

Velocity of mirror = 2 m/s in the opposite direction = -2 m/s

Therefore Velocity of Man w.r.t mirror = V ( man ) - V ( mirror )

⇒ Velocity of Man w.r.t mirror = 5 - ( - 2 )  = 7 m/s

Therefore Velocity of image w.r.t mirror = -7 m/s

Velocity of image w.r.t ground = V( img/mirror ) + V ( mirror )

⇒ Velocity of image w.r.t ground = -7 m/s - 2 m/s = -9 m/s

[If you find any errors, please inbox regarding the error.]

Answer 2

❶

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{\theta}\:=\:m}$

Where,

• θ = 45°

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{45^{\circ}}}$

$:\longrightarrow\:\:\bf{8}$

Thus,

➻ m = 8, which is an even number.

Hence,

➻ No. of images (n) = m - 1

➻ No. of images (n) = 8 - 1

➻ No. of images (n) = 7

$\Large\bf{Therefore,}$

(b) Number of images formed are 7.

_____________________

❷

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{\theta}\:=\:m}$

Where,

• θ = 60°

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{60^{\circ}}}$

$:\longrightarrow\:\:\bf{6}$

Thus,

➻ m = 6, which is an even number.

Hence,

➻ No. of images (n) = m - 1

➻ No. of images (n) = 6 - 1

➻ No. of images (n) = 5

$\Large\bf{Therefore,}$

(a) Number of images formed are 5.

_____________________

❸

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{\theta}\:=\:m}$

Where,

• θ = 120°

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{120^{\circ}}}$

$:\longrightarrow\:\:\bf{3}$

Thus,

➻ m = 3, which is an odd number.

Hence,

➻ No. of images (n) = m

➻ No. of images (n) = 3

$\Large\bf{Therefore,}$

(b) Number of images formed are 3.

_____________________

❹

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{\theta}\:=\:m}$

Where,

• θ = 90°

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{90^{\circ}}}$

$:\longrightarrow\:\:\bf{4}$

Thus,

➻ m = 4, which is an even number.

Hence,

➻ No. of images (n) = m - 1

➻ No. of images (n) = 4 - 1

➻ No. of images (n) = 3

$\Large\bf{Therefore,}$

(a) Number of images formed are 3.

_____________________

❺

↝ All the mirrored walls are perpendicular to each other, i.e

• θ = 90°

★ Considering two adjacent walls.

$:\longrightarrow\:\:\bf{\dfrac{360^{\circ}}{90^{\circ}}}$

$:\longrightarrow\:\:\bf{4}$

Thus,

➻ m = 4, which is an even number.

Hence,

➻ No. of images (n) = m - 1

➻ No. of images (n) = 4 - 1

➻ No. of images (n) = 3

↝ These images formed by the combination of two adjacent walls will be in addition to the object itself act objects for the floor mirror (totally 4), thus the total number of images formed will be,

➻ No. of images (n') = 3 + 4

➻ No. of images (n') = 7

$\Large\bf{Therefore,}$

Number of images formed are 7.

_____________________

❻

Given,

• Velocity of a man towards a stationary mirror is 5 m/s.

✯ Speed of image :

➣ Speed of object = Speed of image

➣ Speed of image = 5 m/s

Since,

↝ The direction of movement is opposite, thus

➣ Speed of image = - 5 m/s

$\Large\bf{Therefore,}$

(c) The velocity of the image w.r.t ground is - 5 m/s.

_____________________

❼

Given,

• Speed of a man away from a stationary mirror is - 10 m/s (negative direction).

✯ Speed of image :

➣ Speed of object = Speed of image

➣ Speed of image = - 10 m/s

Since,

↝ The direction of movement is opposite, thus

➣ Speed of image = - (- 10) m/s

➣ Speed of image = 10 m/s

✯ Image approaching the running man with same speed. Hence relative velocity of image w.r.t running man is

➣ Relative velocity of image = 10 - (-10)

➣ Relative velocity of image = 20 m/s

$\Large\bf{Therefore,}$

The velocity of the image w.r.t man is 20 m/s.

_____________________

❽

Given,

• Speed of a mirror towards a stationary man is 5 m/s.

↝ The speed of the image will be twice that of the object i.e. 10 m/s.

➣ This is due to the fact that when you bring the mirror closer to the object, the image also moves closer.

➣ Thus when we move the mirror closer by 5 metres in one second, the image gets closer by 10 metres.

➣ Therefore the image covers twice the distance in same time interval and has its speed equal to twice the object's.

$\Large\bf{Therefore,}$

(b) The speed of the image w.r.t ground is 10 m/s.

_____________________

❾

Given,

• Speed of a mirror away from a stationary man is 6 m/s.

➣ When we move the mirror away by 6 metres in one second, the image gets away by 12 metres.

Since,

↝ The mirror is moving away from the man, it's value is opposite.

$\Large\bf{Therefore,}$

(b) The speed of the image w.r.t ground is -12 m/s.

_____________________

❿

Given,

• Speed of a man towards a mirror 5 m/s.

• Speed of mirror towards the man is 2 m/s.

[NOTE - Man is moving opposite, according to the movement of the mirror, thus Speed of the man is - 5 m/s.]

➻ Speed of image w.r.t mirror = (Speed of object) - (Speed of mirror)

➻ Speed of image w.r.t mirror = (-5) - (2)

➻ Speed of image w.r.t mirror = - 5 - 2

➻ Speed of image w.r.t mirror = - 7 m/s

Thus,

➙ Speed of image w.r.t mirror = (Speed of image w.r.t mirror) - (Speed of mirror)

➙ Speed of image w.r.t mirror = - 7 - 2

➙ Speed of image w.r.t mirror = - 9 m/s

$\Large\bf{Therefore,}$

The speed of image w.r.t ground is - 9 m/s.